Question

Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.60.

You have in front of you

100 mL of 7.00×10⁻²M HCl,

100 mL of 5.00×10⁻²M NaOH, and

plenty of distilled water.

You start to add HCl to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH. Once you realize your error, you assess the situation. You have 84.0 mL of HCl and 87.0 mL of NaOH left in their original containers.

Assuming the final solution will be diluted to 1.00 L , how much more HCl should you add to achieve the desired pH?

Answer 1

Firstly we determine HCl is in the solution.

100mL(Amount started with) - 84mL(amount left) = 16mL of HCl

Similarly, amount of NaOH in the solution would be

100mL(amount started with) - 87mL(amount left) = 13mL of NaOH

Converting volume into moles,

(16ml/1000)L x 7.00*10^-2M = 1.12*10^-3mol of HCl

(13mL/1000)L x 5.00*10^-2M = 6.5*10^-4mol of NaOH

Now, HCl will have reacted with NaOH. After reacting, an amount of HCl will be neutralized by the NaOH. Lets calculate the remaining HCl. Because 1 mole of HCL reacts with 1 mole of NaOH,

1.12*10^-3mol HCl - 6.5*10^-4mol NaOH = 4.70*10^-4 mol HCl

The total amount of HCl needed to reach a pH of 2.70 would be

pH = -log [H+]

2.60 = -log[H+]

10^-(2.60) = [H+]

2.512*10^-3 = [H+]

Converting this concentration in to moles.

Molarity = Moles/Liters

Molarity x Liters = Moles

2.512*10^-3M x 1.00L = 2.512*10^-3 mol HCl is needed to have a pH of 2.60

Now, the amount of HCL that needs to be added to the solution would be

(Total moles of HCl needed) - (Moles of HCl in solution) = Moles needed to be added

2.512*10^-3mol - 4.70*10^-4mol = 2.042*10^-3 mol HCl

Convert Moles of HCl to mL of HCl.

Molarity = Moles/Liter

Moles/Molarity = Liter

(2.042*10^-3mol) / (7.00*10^-2M) = 2.917*10^-2 L = 29.17mL HCl